What is the name of a molecule composed of one or more sugars?

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In t

const2013 [10]

Answer:

$\boxed{\text{47.4 g}}$

Explanation:

We are given the mass of two reactants, so this is a limiting reactant problem.

We know that we will need mases, moles, and molar masses, so, let's assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r: 17.03 32.00 18.02

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

m/g: 70.1 70.1

Step 1. Calculate the moles of each reactant

$\text{Moles of CO } = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{17.03 g}} = \text{4.116 mol}\\\\\text{Moles of H$_{2}$O} = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{32.00 g}} = \text{2.191 mol}$

Step 2. Identify the limiting reactant

Calculate the moles of H₂O we can obtain from each reactant.

From NH₃:

The molar ratio of H₂O:NH₃ is 6:4.

$\text{Moles of H$_{2}$O} = \text{4.116 mol NH$_{3}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{4 mol NH$_{3}$}} = \text{6.174 mol H$_{2}$O}$

From O₂:

The molar ratio of H₂O:O₂ is 6:5.

$\text{Moles of H$_{2}$O} = \text{2.191 mol O$_{2}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{5 mol O$_{2}$}} = \text{2.629 mol H$_{2}$O}$

O₂ is the limiting reactant because it gives the smaller amount of H₂O.

Step 3. Calculate the theoretical yield.

$\text{Theor. yield } = \text{2.629 mol H$_{2}$O}\times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{47.4 g H$_{2}$O}\\\\\text{The maximum yield of H$_{2}$O is }\boxed{\textbf{47.4 g}}$

6 0

3 years ago

The ksp of manganese(ii) carbonate, mnco3, is 2.42 × 10-11. calculate the solubility of this compound in g/l.

Vikki [24]

Answer : The solubility of this compound in g/L is $565.414\times 10^{-6}g/L$ .

Solution : Given,

$K_{sp}=2.42\times 10^{-11}$

Molar mass of $MnCO_3$ = 114.945g/mole

The balanced equilibrium reaction is,

$MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3$

At equilibrium s s

The expression for solubility constant is,

$K_{sp}=[Mn^{2+}][CO^{2-}_3]$

Now put the given values in this expression, we get

$2.42\times 10^{-11}=(s)(s)\\2.42\times 10^{-11}=s^2\\s=0.4919\times 10^{-5}=4.919\times 10^{-6}moles/L$

The value of 's' is the molar concentration of manganese ion and carbonate ion.

Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.

$s=4.919\times 10^{-6}moles/L\times 114.945g/mole=565.414\times 10^{-6}g/L$

Therefore, the solubility of this compound in g/L is $565.414\times 10^{-6}g/L$ .

8 0

3 years ago

Una solución formada por 58 gramos de cloruro de sodio (NaCl) en 100 mL de solución posee una concentración

masya89 [10]

Answer:

58

Explanation:

8 0

3 years ago

Identify the solution that contains both the solute and solvent in the solid phase. A) air B) NaCl C) bronze D) salt water

Iteru [2.4K]

Answer:prolly c

Explanation:

5 0

3 years ago

Reaction intermediates differ from activated complexes in that A. they are stable molecules with normal bonds and are frequently

vova2212 [387]

Answer:

they are molecules with normal bonds rather than partial bonds and can occasionally be isolated.

Explanation:

In chemistry, reaction intermediates are species that are formed from reactants and are subsequently being transformed into products as the reaction progresses. In other words, reaction intermediates are species that do not appear in a balanced reaction equation but occur somewhere along the reaction mechanism of a non-elementary reaction. They are usually short lived species that possess a high amount of energy. They may or may not be isolated.

They are often molecular species with normal bonds unlike activated complexes that are sometimes hypervalent species.

3 0

3 years ago